Problem: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $115$ years; the standard deviation is $13.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living longer than $87.8$ years.
Solution: $115$ $101.4$ $128.6$ $87.8$ $142.2$ $74.2$ $155.8$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $115$ years. We know the standard deviation is $13.6$ years, so one standard deviation below the mean is $101.4$ years and one standard deviation above the mean is $128.6$ years. Two standard deviations below the mean is $87.8$ years and two standard deviations above the mean is $142.2$ years. Three standard deviations below the mean is $74.2$ years and three standard deviations above the mean is $155.8$ years. We are interested in the probability of a turtle living longer than $87.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the turtles will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $87.8$ years and the other half $({2.5\%})$ will live longer than $142.2$ years. The probability of a particular turtle living longer than $87.8$ years is ${95\%} + {2.5\%}$, or $97.5\%$.